YES(O(1),O(n^2))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , sum1(0()) -> 0()
  , sum1(s(x)) -> s(+(sum1(x), +(x, x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.

Trs:
  { sum(s(x)) -> +(sum(x), s(x))
  , sum1(s(x)) -> s(+(sum1(x), +(x, x))) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
    [sum](x1) = 3*x1 + x1^2
                           
        [0]() = 0          
                           
      [s](x1) = 1 + x1     
                           
  [+](x1, x2) = x1 + x2    
                           
   [sum1](x1) = x1 + x1^2  
                           
  
  This order satisfies the following ordering constraints.
  
      [sum(0())] =                          
                 >=                         
                 =  [0()]                   
                                            
     [sum(s(x))] =  4 + 5*x + x^2           
                 >  4*x + x^2 + 1           
                 =  [+(sum(x), s(x))]       
                                            
     [sum1(0())] =                          
                 >=                         
                 =  [0()]                   
                                            
    [sum1(s(x))] =  2 + 3*x + x^2           
                 >  1 + 3*x + x^2           
                 =  [s(+(sum1(x), +(x, x)))]
                                            

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict Trs:
  { sum(0()) -> 0()
  , sum1(0()) -> 0() }
Weak Trs:
  { sum(s(x)) -> +(sum(x), s(x))
  , sum1(s(x)) -> s(+(sum1(x), +(x, x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'custom shape polynomial interpretation' to
orient following rules strictly.

Trs:
  { sum(0()) -> 0()
  , sum1(0()) -> 0() }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^2)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-restricted polynomial
  interpretation.
    [sum](x1) = 3*x1 + x1^2
                           
        [0]() = 1          
                           
      [s](x1) = 1 + x1     
                           
  [+](x1, x2) = x1 + x2    
                           
   [sum1](x1) = x1 + 2*x1^2
                           
  
  This order satisfies the following ordering constraints.
  
      [sum(0())] = 4                       
                 > 1                       
                 = [0()]                   
                                           
     [sum(s(x))] = 4 + 5*x + x^2           
                 > 4*x + x^2 + 1           
                 = [+(sum(x), s(x))]       
                                           
     [sum1(0())] = 3                       
                 > 1                       
                 = [0()]                   
                                           
    [sum1(s(x))] = 3 + 5*x + 2*x^2         
                 > 1 + 3*x + 2*x^2         
                 = [s(+(sum1(x), +(x, x)))]
                                           

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

Weak Trs:
  { sum(0()) -> 0()
  , sum(s(x)) -> +(sum(x), s(x))
  , sum1(0()) -> 0()
  , sum1(s(x)) -> s(+(sum1(x), +(x, x))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

Hurray, we answered YES(O(1),O(n^2))